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Vectorization

Adapted from Software Carpentry

Overview

Questions:

How can I operate on all the elements of a vector at once?

Objectives:

To understand vectorized operations in R

Introduction

Most of R’s functions are vectorized, meaning that the function will operate on all elements of a vector without needing to loop through and act on each element one at a time. This makes writing code more concise, easy to read, and less error prone.

x <- 1:4
x * 2

[1] 2 4 6 8

The multiplication happened to each element of the vector.

We can also add two vectors together:

y <- 6:9
x + y

[1]  7  9 11 13

Each element of x was added to its corresponding element of y:

 x:  1  2  3  4
    +  +  +  +
 y:  6  7  8  9
---------------
    7  9 11 13

Vectorized Operations vs. Loops

Here is how we would add two vectors together using a for loop:

output_vector <- c()
for (i in 1:4) {
  output_vector[i] <- x[i] + y[i]
}
output_vector

[1]  7  9 11 13

Compare this to the output using vectorised operations:

sum_xy <- x + y
sum_xy

[1]  7  9 11 13

The vectorized version is much more concise and easier to read!

Challenge 1

Let’s try this on the pop column of the gapminder dataset.

Make a new column in the gapminder data frame that contains population in units of millions of people. Check the head or tail of the data frame to make sure it worked.

Solution to Challenge 1

gapminder$pop_millions <- gapminder$pop / 1e6
head(gapminder)

The pop_millions column now contains the population in millions.

Challenge 2

On a single graph, plot population, in millions, against year, for all countries. Do not worry about identifying which country is which.

Repeat the exercise, graphing only for China, India, and Indonesia. Again, do not worry about which is which.

Solution to Challenge 2

# All countries
ggplot(gapminder, aes(x = year, y = pop_millions)) +
  geom_point()

# China, India, and Indonesia
countryset <- c("China", "India", "Indonesia")
ggplot(gapminder[gapminder$country %in% countryset, ],
       aes(x = year, y = pop_millions)) +
  geom_point()

Vectorized Operations with Comparisons and Logic

Comparison operators, logical operators, and many functions are also vectorized:

Comparison operators:

x > 2

[1] FALSE FALSE  TRUE  TRUE

Logical operators:

a <- x > 3  # or, for clarity, a <- (x > 3)
a

[1] FALSE FALSE FALSE  TRUE

Tip: Some Useful Functions for Logical Vectors

any() will return TRUE if any element of a vector is TRUE.

all() will return TRUE if all elements of a vector are TRUE.

Vectorized Functions

Most functions also operate element-wise on vectors:

x <- 1:4
log(x)

[1] 0.0000000 0.6931472 1.0986123 1.3862944

Vectorized Operations on Matrices

Vectorized operations work element-wise on matrices:

m <- matrix(1:12, nrow = 3, ncol = 4)
m * -1

     [,1] [,2] [,3] [,4]
[1,]   -1   -4   -7  -10
[2,]   -2   -5   -8  -11
[3,]   -3   -6   -9  -12

Tip: Element-wise vs. Matrix Multiplication

Very important: the operator * gives you element-wise multiplication!

To do matrix multiplication, we need to use the %*% operator:

m %*% matrix(1, nrow = 4, ncol = 1)

     [,1]
[1,]   22
[2,]   26
[3,]   30

matrix(1:4, nrow = 1) %*% matrix(1:4, ncol = 1)

     [,1]
[1,]   30

For more on matrix algebra, see the Quick-R reference guide.

Challenge 3

Given the following matrix:

m <- matrix(1:12, nrow = 3, ncol = 4)
m

     [,1] [,2] [,3] [,4]
[1,]    1    4    7   10
[2,]    2    5    8   11
[3,]    3    6    9   12

Write down what you think will happen when you run:

m ^ -1
m * c(1, 0, -1)
m > c(0, 20)
m * c(1, 0, -1, 2)

Did you get the output you expected? If not, ask a helper!

Solution to Challenge 3

m ^ -1

          [,1]      [,2]      [,3]       [,4]
[1,] 1.0000000 0.2500000 0.1428571 0.10000000
[2,] 0.5000000 0.2000000 0.1250000 0.09090909
[3,] 0.3333333 0.1666667 0.1111111 0.08333333

m * c(1, 0, -1)

     [,1] [,2] [,3] [,4]
[1,]    1    4    7   10
[2,]    0    0    0    0
[3,]   -3   -6   -9  -12

m > c(0, 20)

      [,1]  [,2]  [,3]  [,4]
[1,]  TRUE FALSE  TRUE FALSE
[2,] FALSE  TRUE FALSE  TRUE
[3,]  TRUE FALSE  TRUE FALSE

m * c(1, 0, -1, 2)

     [,1] [,2] [,3] [,4]
[1,]    1    4    7   10
[2,]    0    0    0    0
[3,]   -3   -6   -9  -12
[4,]    8   10   12   14

Note how the shorter vector was recycled to match the dimensions of the matrix!

Challenge 4

We’re interested in looking at the sum of the following sequence of fractions:

x = 1/(1^2) + 1/(2^2) + 1/(3^2) + ... + 1/(n^2)

This would be tedious to type out, and impossible for high values of n. Use vectorisation to compute x when n=100. What is the sum when n=10,000?

Solution to Challenge 4

sum(1/(1:100)^2)

[1] 1.634984

sum(1/(1:10000)^2)

[1] 1.644834

This converges to \(\pi^2/6 \approx 1.645\).

Recycling: Operations on Vectors of Unequal Length

Tip: Operations on Vectors of Unequal Length

Operations can also be performed on vectors of unequal length, through a process known as recycling. This process automatically repeats the smaller vector until it matches the length of the larger vector. R will provide a warning if the larger vector is not a multiple of the smaller vector.

x <- c(1, 2, 3)
y <- c(1, 2, 3, 4, 5, 6, 7)
x + y

Warning in x + y: longer object length is not a multiple of shorter object
length

[1] 2 4 6 5 7 9 8

Vector x was recycled to match the length of vector y:

 x:  1  2  3  1  2  3  1
    +  +  +  +  +  +  +
 y:  1  2  3  4  5  6  7
-----------------------
    2  4  6  5  7  9  8

Key Points

Use vectorized operations instead of loops
Vectorized code is more concise, easier to read, and less error-prone
Most R functions operate element-wise on vectors
Be aware of vector recycling when operating on vectors of different lengths
Use %*% for matrix multiplication, not *